Simplify the following expression: $y = \dfrac{-9x^2- 40x- 16}{-9x - 4}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-16)} &=& 144 \\ {a} + {b} &=& &=& {-40} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $144$ and add them together. The factors that add up to ${-40}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-4}$ and ${b}$ is ${-36}$ $ \begin{eqnarray} {ab} &=& ({-4})({-36}) &=& 144 \\ {a} + {b} &=& {-4} + {-36} &=& -40 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 {-4}x) + ({-36}x {-16}) $ Factor out the common factors: $ x(-9x - 4) + 4(-9x - 4)$ Now factor out $(-9x - 4)$ $ (-9x - 4)(x + 4)$ The original expression can therefore be written: $ \dfrac{(-9x - 4)(x + 4)}{-9x - 4}$ We are dividing by $-9x - 4$ , so $-9x - 4 \neq 0$ Therefore, $x \neq -\frac{4}{9}$ This leaves us with $x + 4; x \neq -\frac{4}{9}$.